MCB142/IB163 Mendelian and Population Genetics 9/19/02

MCB142/IB163 Mendelian and Population Genetics 9/19/02 Practice questions lectures 5-12 (Hardy Weinberg, chi-square test, Mendel s second law, gene interactions, linkage maps, mapping human diseases, non-random mating, selection, genetic drift, evolutionary genetics) Question 1 You are studying an X-linked trait. There are two alleles, one showing complete dominance over the other. In females, 84 percent show the dominant phenotype. What percentage of the males will show the dominant phenotype, assuming that individuals in this large population are in Hardy Weinberg proportions, and that the allele frequencies are equal in males and females? Question 2 Human albinism is an autosomal recessive trait. Suppose that you find a village in the Andes where 1/4 of the population is albino. If the population size is 1000 and the population is in Hardy-Weinberg proportions with respect to this trait, how many individuals are expected to be heterozygotes? Question 3 An RFLP probe shows four different morphs (alleles) in a population of three-spined sticklebacks (a fish-gasterosteus aculeatus). How many different heterozygous genotypes will there be in this population with regard to this RFLP? Question 4 Suppose that you are a genetic counselor, and a couple seeks your advice about BHT tasting (people who can taste BHT in processed food are recessive homozygotes for the taster allele and non-tasters are homozygous dominant or heterozygous for the non-taster allele). Both prospective parents are non-tasters, but a careful analysis of the husband s pedigree reveals that he is a carrier of the taster allele. The proportion of BHT tasters in the population is 16 percent. What is the probability that the couple s first child will be a non-taster of BHT assuming the population is in Hardy Weinberg proportions? Question 5 In a population under study, the frequency of an autosomal recessive disease is 1 percent. Assuming Hardy Weinberg proportions, what is the probability of two heterozygotes marrying, and what proportion of affected children will be born from such matings? Question 6 In a Pygmy group in central Africa, the frequencies of alleles determining the ABO blood groups were estimated as 0.2 for I A, 0.1 for I B, and 0.7 for i (allele O). Assuming Hardy Weinberg proportions, determine the expected frequencies of the blood types A, B, AB, and O. (Alleles A and B are codominant, and O is recessive.) Page 1

Question 7 In a test of Hardy Weinberg proportions for a gene with 3 alleles, and hence 6 genotype classes, determine whether the chi-square value given below for a number of different cases is compatible with the null hypothesis, or rejects the null hypothesis of Hardy Weinberg proportions, in which case give the P value. You will need to determine the degrees of freedom (df) in each case. (a) The chi-square value is 12.5 and all expected values for the 6 genotype classes are >5. (b) The chi-square value is 4.3, one genotype class with expected value <5 was combined with another, for a total of 5 genotype classes. (c) The chi-square value is 20.5, two genotype classes with expected values <5 were combined together (the expected value was then >5). (d) Describe what the P value means in each of the cases where the null hypothesis was rejected. Question 8 A plant of genotype C/C ; d/d is crossed to c/c ; D/D and an F 1 testcrossed to c/c ; d/d. (a) If the genes are on separate chromosomes, what percentage of c/c d/d offspring will be seen? (b) If the genes are linked, and 20 map units apart, what percentage of c/c d/d offspring will be seen? Question 9 In Drosophila, the two genes w and sn are X-linked and 25 map units apart. A female fly of genotype w + sn + /w sn is crossed to a male from a wild-type line. What percent of male progeny will be w + sn? Question 10 Three genes, each represented by a normal (dominant) and a mutant (recessive) allele, are on separate chromosomes and assort independently of one another. The F 2 of a cross of the homozygous (A/A ; B/B ; C/C) normal strain with the triply mutant strain (a/a ; b/b ; c/c) will produce 3/4 normal to 1/4 mutant for each character. Using these figures, and without a checkerboard (Punnett square), determine what fractions of the F 2 will be of the phenotypes indicated below. Show the component probabilities. (a) A/- ; B/- ; c/c (b) a/a ; b/b ; c/c (c) A/- ; B/- ; C/- (d) a/a ; B/- ; c/c Question 11 A female fruit fly of genotype A/a ; B/b is crossed to a male fruit fly of genotype a/a ; b/b. The progeny of this cross were Genotype Number of individuals A/a ; B/b 38 a/a ; b/b 37 A/a ; b/b 12 a/a ; B/b 13 (a) What gametes were produced by parent 1 and in what proportions? (b) Do these proportions represent independent assortment of the two genes? (c) What can be deduced from these proportions? (d) Draw a diagram or explain the origin of the two rarest progeny genotypes. Page 2

Question 12 In mice the allele for color expression is C (c = albino). Another gene determines color (B = black and b = brown). Yet another gene modifies the amount of color so that D = normal amount of color and d = dilute (milky) color. Two mice that are C/c ; B/b ; D/d are mated. What proportion of progeny will be dilute brown (assume complete dominance at each locus and independent assortment of the genes)? Question 13 In pigs, when a pure-breeding red is crossed to a pure-breeding white the F 1 are all red. However, the F 2 shows 9/16 red, 1/16 white, and 6/16 are a new color, sandy. (a) Provide a genetic hypothesis for this observation (b) How would you test this hypothesis? Question 14 The genes Dm/dm, Lu/lu, and Sc/sc are inherited autosomally in humans. In a large population, you find women who are phenotypically dominant for all three traits whose parents consisted of one phenotypically dominant individual (for all three traits) and one phenotypically recessive individual (for all three traits). You observe the progeny of all such women whose husbands have a recessive phenotype for these three traits, and obtain the following data Phenotypes Number of individuals Dm Lu Sc 407 Dm Lu sc 4 Dm lu sc 22 Dm lu Sc 72 dm Lu Sc 18 dm lu Sc 3 dm Lu sc 70 dm lu sc 404 1,000 (a) What is the map order of the three loci? (b) What are the map distances between the three markers? Question 15 Most genetic variation for many human loci lies within local populations rather than between populations or races. What does this observation tell you about human genetic evolution? Question 16 Most genetic variation for many human loci lies within local populations rather than between populations or races. There are some contradictions to this general trend, such as hair and skin color, and sickle cell anemia. What do these exceptions tell you about the evolution of those traits? Question 17 Compare and contrast the neutral, balance, and evolutionary lag schools. Page 3

Question 18 For each of the following sets of genotype relative fitness values, predict whether a stable polymorphism will be maintained (in which case give the equilibrium frequency of the two alleles) or whether one or the other of the alternate alleles will go to fixation (in which case state which allele will become fixed in the population). Assume that the population size is very large and that selection is the only force acting to alter the frequencies of the alleles at this locus. AA AB BB (a) 1.00 0.98 0.10 (b) 0.80 1.00 0.6 (c) 1.0 0.6 0.6 Question 19 At a locus with two alleles A and B, the homozygous AA individual has a fitness of 0.5 relative to the heterozygote AB, while the BB genotype is lethal. Under this selection scheme (a) what is the equilibrium allele frequency of A? (b) what are the three genotype frequencies expected at zygote formation, assuming random mating? Question 20 What will determine the rate of a molecular clock for a gene that codes for a functional protein? Question 21 To what extent is evolution a chance process and to what extent a directed process? Question 22 In the thousands of years that dogs have been domesticated, breeders have developed numerous breeds with many desirable traits. In many breeds of dogs, however, undesirable characteristics have appeared as well. The undesirable features include single gene traits such as retinal degeneration or deafness. Why do these undesirable traits persist? Page 4

MCB142/IB163 Mendelian and Population Genetics 9/19/02 Answers to Practice questions lectures 5-12 Question 1 Let allele A represent the dominant allele, with frequency p in both males and females, and a the recessive allele, with frequency q, with p + q =1. We are given the information that in females 84% have the dominant phenotype. This means that 16% have the recessive phenotype aa, with expected frequency q 2, so using the square root formula (we can t use allele counting since we cannot distinguish the heterozygotes from the dominant homozygote class) under the assumption that the population is in Hardy Weinberg proportions, the estimate of q = 0.4, hence p = 0.6. In males, the frequency of the dominant phenotype is equal to the allele frequency, hence the answer is 0.6. Question 2 We represent the autosomal recessive allele for albinism by a, with frequency q, and the dominant allele is represented by A, with frequency p, with p + q =1. We are told to assume Hardy Weinberg proportions, thus we can use the square root formula to estimate q. Since q 2 = 0.25, the estimate of q = 0.5. The expected proportion of heterozygous individuals is 2pq = 0.5, thus in the population of size 1,000 we expect 500 individuals to be heterozygotes, and hence carriers of the disease predisposing allele. Question 3 Denoting the four alleles as A, B, C, and D, then there are 6 heterozygous genotypes, AB, AC, AD, BC, BD, and CD. (There are 4 homozygous genotypes possible: AA, BB, CC, and DD.) Question 4 Denote the recessive taster allele by t (frequency q), so that tasters are tt (frequency q 2 ). We are given that the husband is heterozygous, hence Tt. We are given that the mother is a non-taster, so she is either TT or Tt. Assuming Hardy Weinberg proportions, we estimate q using the square root formula: given that q 2 = 0.16, then the estimate is q = 0.4, hence p = 0.6. The frequency of homozygous TT individuals in the population is p 2 = 0.36, and the frequency of heterozygotes Tt is 2pq = 2(0.6)(0.4) = 0.48. Given that the woman is a non-taster, i.e., she is either TT or Tt, the conditional probability that she is TT is 0.36/0.84 = 0.4286, and the probability she is Tt, is 0.48/ 0.84 = 0.5714. We have to consider both possible genotypes of the mother, since a non-taster child is possible both if she is TT or Tt. Thus the required probability is 1 (the probability the man is Tt) x 0.4286 (the probability the woman is TT given she is a non-taster) x 1 (the probability that their first child will be a non-taster (TT or Tt), given these genotypes) (= 0.4286) + 1 (the probability the man is Tt) x 0.5714 (the probability the woman is Tt given she is a non-taster) x 3/4 (the probability that their first child will be a nontaster (TT or Tt) given these genotypes) (= 0.4286), thus the required probability is 0.8572 (= 0.4286 + 0.4286, note that it is a coincidence of the given allele frequencies that the probabilities from both possible genotypes of the woman are the same, usually these would be different). A quicker way to get the answer, is to calculate the probability their first child will be a taster = 1 (the probability the man is Tt) x 0.5714 (the probability the woman is Tt given she is a non-taster) x 1/4 (the probability that their first child will be a taster (tt), given these genotypes = 0.1428, and thus the required probability is 1 0.1428 = 0.8572. Page 5

Question 5 Assuming Hardy Weinberg proportions, we use the square root formula to calculate the allele frequency q of the recessive allele, denoted by a: given q 2 =0.01, q = 0.1, hence p = 0.9. The number of heterozygous carriers (Aa) is thus 2pq = 0.18, and the frequency of heterozygous x heterozygous matings (assuming mating is at random) is (2pq) x (2pq) = 0.0324. Since ¼ of the children from such a mating will be affected (frequency = 0.0324/4 = 0.0081), this represents 81% of the total cases which overall have a frequency of 1%. Question 6 f(a) = (0.2) 2 + 2(0.2)(0.7) = 0.32 f(b) = = (0.1) 2 + 2(0.1)(0.7) = 0.15 f(ab) = 2(0.2)(0.1) = 0.04 f(oo) = (0.7) 2 = 0.49 (check that the sum of the genotype frequencies adds to 1) Question 7 (a) df = 3 (=6 1 2, 6 classes minus 1 since we know the total count, minus 2 allele frequency estimates, the third allele frequency is known once we know 2 of the allele frequencies), the chisquare value of 12.5 is between the P<0.01 and P<0.001 values for a chi-square distribution with 3 df (see table from handout), so we reject the null hypothesis with P<0.01. (b) df = 2 (5 1 2), the chi-square value of 4.3 with 2 df is compatible with the null hypothesis, so these data are not significant. (c) df = 2 (5 1 2), the chi-square of 20.5 with 2 df is very significantly different from the null hypothesis (fit to Hardy Weinberg proportions) with P<0.001. (d) The P value represents the type 1 error, which is the probabaility that you have rejected the null hypothesis when it was true, i.e., the probability you have made a mistake in rejecting the null hypothesis. So in (a) the type 1 error is 0.01, while in (c) it is smaller at 0.001. Question 8 (a) All F1 individuals will be C/c D/d and will produce equal numbers (since the genes are on different chromosomes) of the 4 gametic types CD, Cd, cd, and cd. Since the testcross individual only produces gametes of type cd, the proportion of c/c d/d offspring will be 25%. (b) All F1 individuals will again be C/c D/d. However, since the genes are now linked (r = 20 cm = 20% recombination) apart, and we know the parental gametic types are Cd and cd, the 4 gametic types CD, Cd, cd, and cd are expected in proportions r/2, (1-r)/2, (1-r)/2, and r/2 respectively, where r = 0.2. Since the testcross individual only produces gametes of type cd, the proportion of c/c d/d offspring will be 10%. Question 9 The female will produce four gametes: the 2 parental w + sn + and w sn will occur with equal frequencies of (1-r)/2 where r = 0.25, i.e., each with 37.5% frequency, while the 2 non-parental (recombinant) gametes w + sn and w s + will each occur with a frequency of 12.5% (check total number of gametes = 100%). Since males have only one X chromosome which they receive from their mothers, the percent of male progeny that will be w + sn is 12.5%. Question 10 (a) ¾ x ¾ x ¼ = 9/64 (b) ¼ x ¼ x ¼ = 1/64 (c) ¾ x ¾ x ¾ = 27/64 (d) ¼ x ¾ x ¼ = 3/64 Page 6

Question 11 (a) The proportions were 38% AB, 37% ab, 12% Ab, and 13% ab. (b) No, because we expect 25% of each under independent assortment, and these numbers are very different from that. (One could do the chi-square test to show lack of independent assortment, but it is not asked for in this question.) (c) The two genes must be linked; the %R (RF) = (12 + 13) / 100 = 25%, i.e., the two loci are 25 map units (25 cm) apart on the same chromosome. (d) The female parent received gametes AB and ab from her two parents (she is AB/ab), the male parent received ab gametes from both his parents (he is ab/ab), and the two rarest progeny classes represent recombinant classes from the female parent. Question 12 We need to determine the probability of a C/- ; bb ; dd offspring from this cross = ¾ x ¼ x ¼ = 3/64. Question 13 (a) The presence of a dominant allele at either of two loci (A/- ; b/b or a/a ; B/-) gives sandy, where red is determined by the dominant alleles of both loci (A/- ; B/-), and white by the recessive genotype at both loci (a/a ; b/b). (b) Do testcrosses of a large number of the sandy offspring with the white offspring. Under this hypothesis, the sandy offspring are A/A ; b/b (1/16) (gametes Ab) or A/a ; b/b (2/16) (gametes ½ Ab and ½ ab) or a/a ; B/B (1/16) (gametes ab) or a/a ; B/b (2/16) (gametes ½ ab and ½ ab), i.e., three types of gametes with equal frequencies: Ab, ab, and ab, so the offspring from the testcrosses with (a/a ; b/b) should be equal proportions of A/a ; b/b (sandy) a/a ; B/b (sandy) a/a ; b/b (white), i.e., 2/3 sandy and 1/3 white. (One should also perform other crosses to confirm the model, including testcrosses and crosses of the putative genotypes of specific sandy offspring.) Question 14 (a) The non-recombinant gametes from the mother are Dm Lu Sc and dm lu sc, while the double recombinant gametes (the least frequent class) are Dm Lu sc and dm lu Sc, thus the order of the three loci is Dm Sc Lu or Lu Sc Dm. (b) The Dm Sc recombination distance is (4 + 22 + 18 + 3) / 1,000 = 4.7% (= 4.7 mu = 4.7 cm). The Sc Lu recombination distance is (4 + 70 + 72 + 3) / 1,000 = 14.9%. The Dm Sc recombination distance is (22 + 72 + 18 + 70 + 4 x 2 + 3 x 2) / 1,000 = 19.6%. Correct answer must include double recombinants for Dm Sc, and check that the recombination distances are consistent, i.e., 4.7 + 14.9 = 19.6. Question 15 It suggests that humans have had persistent migration between populations that has prevented localized differentiation for most genes Page 7

Question 16 They most likely reflect adaptations to local conditions. Dark skin is more adaptive in warm sunny climates, whereas light skin may be more adaptive when exposure to sun is minimal. The sickle cell anemia allele provides protection against malaria (in heterozygotes and homozygotes), and so is at high frequency in malarial environments, despite the deleterious effects of the allele in homozygous individuals with sickle cell anemia. Other genes which do not contribute to local adaptation would not show differentiation (the majority of genes). Question 17 The neutral school maintains that much of the genetic variation in populations represents allelic variants that are selectively equivalent to each other and polymorphisms seen in a population represent a balance of new mutations entering the population, some of which will replace older alleles by drift (chance) effects. In this way genetic changes accumulate between distinct species. The balance school maintains that much of the genetic variation in populations has adaptive significance. The evolutionary lag school argues that much of the variation seen in populations may represent transient variation, as selectively advantageous alleles replace older alleles in a race to keep up with changes in the environment. It is difficult, if not impossible, to know what fraction of the genetic variation in populations falls into each of these three categories, and all we can say is that each contributes to genetic variation. Question 18 (a) A will be fixed. (b) Balanced polymorphism, f(a) = p1 = s2/(s1 + s2) = 0.4/0.6 = 2/3 (s1 = 0.2, s2 = 0.4), f(b) = p2 = s1/(s1 + s2) = 0.2/0.6 = 1/3. (c) A will be fixed. Question 19 (a) Using the formula to determine equilibrium allele frequencies with heterozygote advantage, s1 = 0.5, and s2 = 1.0, so the equilibrium frequency of A = p1 = s2/(s1 + s2) = 1/1.5 = 2/3 (and the equilibrium frequency of B = 0.5/1.5 = 1/3). (b) At zygote formation, the expected genotype frequencies are the Hardy Weinberg proportions 4/9 AA, 4/9 AB, and 1/9 BB. Question 20 The degree to which changes in the protein will decrease the fitness of the organism. If the protein is very sensitive to change (cannot function optimally with even small changes in its amino acid composition) then the molecular clock will be extremely slow. If the protein can function adequately even with many changes, perhaps to sites that are not immediately associated with the active site on the protein, then the molecular clock will be relatively fast. Page 8

Question 21 The ultimate source of genetic variation is mutation, so the building blocks of evolution come from chance events. Also, mutations which are selectively equivalent to existing alleles in the population can become established in the population by genetic drift (chance) effects. Thus, even without direct selection, evolution will occur in populations, since they are always of finite size and subject to drift effects. Evolution is not a directed process in terms of having a purpose or striving for improvement. Since there are many conceivable ways to succeed at surviving and reproducing, natural selection lacks direction in a specific sense. But there are more ways of failing to survive and reproduce than there are to succeed at any point in time, and therefore natural selection provides some direction in moving populations away from traits that fail and towards the traits that succeed Question 22 Development of a new breed of any plant or animal frequently includes inbreeding, and this may expose rare recessive alleles in the gene pool. This is a special problem when breeders start with only a few animals and create entire stocks from their descendants. Correlated responses to selection also contribute to the development of undesirable characteristics when selecting for desirable characteristics. Selection for a nicely shaped head or a particular leg length in a dog may have undesirable effects on other skeletal features. Genetic linkage and association (linkage disequilibrium) of desirable alleles at some loci with undesirable alleles at other loci contribute greatly to the correlated selection of two or more different traits. Page 9